You send a beam of alpha particles of energy 4.522 MeV in toward a very thin pla

Question

You send a beam of alpha particles of energy 4.522 MeV in toward a very thin plastic film. You then detect alpha particles scattering directly backward.

You can accurately measure the energy of the scattered alpha particles. The number of alphas detected as a function of alpha energy is plotted below.

(1.) Why does this spectrum have peaks instead of being a smoothly varying function?

You realize that if you know the energy of each of these peaks, you can use kinematics (use relativistic kinematic here) to give you some clues about the nuclei in the target. You know both total energy and momentum must be conserved in the collision. (Note, you will get the nuclear mass out, not the atomic mass, but the electrons don’t have much mass.)

(2.) Given the energy of the scattered alphas shown in the diagram, determine what nuclei might be present in the target.

*These accurate numbers may help:

Rest energy of an alpha particle: 3727.38 MeV

Rest energy of an electron: 0.510999 MeV

1 amu (1/12 of the mass of a 12C atom) = 931.494 MeV/c2

Explanation / Answer

The aplha particles are elastically scatter by nuclie in the sample and the energy of the back-scattered alpha-particle depends on the mass of the target nuclie. We get the peack energies due to the presence of multiple nuclie in the sample.

incident alpaha energy Ei

scattered alpha energy Es

Mass of alpha Ma

Mass of nucleus Mn

conserving the momentum before and after collision

Ma*vi = Mn *vn + Ma * vf

Ma (vi - vf ) = Mn*vn --------(1)

vi - initial velocity of alpha

vf - alpha velocity after scatter

vn velocity of the nucleus after scatter

It is an eleastic scattering and energy is also conserved

Ma *vi2  = Mn*vn2  + Ma * vf2  

Ma(vi2 -vf2 ) = Mn *vn2   -----(2)

squaring (1)

Ma2 *(vi-vf)2  = Mn2 vn2   ----(3)

divid 3 by 2 we get

Mn = Ma (vi-vf) /(vi +vf)

= Ma * (Mavi2  - Mavf2) / (Mavi2 + Mavf2)

= Ma * (Ei -Ef) /Ei + Ef) = Ma* (4.522 -1.394) /(4.522 +1.394)

= (3727.8 /931.494)* 3.128 /5.916

= 2. 11 u

The nucleus is H-2

energy of the ohter peaks are not given,

The lower energy peak can be H-1 and higher peak that of H-3

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